以下是您希望用显示模式展示的数学公式：

$$
\hat{\beta}_{2} \pm t_{0.025} \times S_{\hat{\beta}_{2}}=0.0286 \pm 2.365 \times 0.0058 \text { 或 }(0.014,0.042)
$$

在 Eviews 软件下, 回归结果如图 3-2 所示:

| - | - | - | - | - | - |
|---|---|---|---|---|---|
| *Equation: UNTITLED* | *Workfi...* | *$-\square x$* |   |   |   |
| View | Procs | Objects | Print Name Freeze | Estimate | Forecast St & Stats & Resids |
| Dependent Variable: $Y$ <br> Method: Least Squares <br> Date: $10 / 21 / 09$ Time: $15: 51$ <br> Sample: 110 <br> Included observations: 10 |   |   |   |   |   |
| Variable | Coefficient | Std. Error | t-Statistic |   | Prob. |
| $c$ <br> $x_{1}$ <br> $x_{2}$ | 626.5093 <br> -9.790570 <br> 0.028618 | 40.13010 <br> 3.197843 <br> 0.005838 | 15.61195 <br> -3.061617 <br> 4.902030 |   | 0.0000 <br> 0.0183 <br> 0.0017 |
| R-squared <br> Adjusted R-squared <br> S.E. of regression <br> Sum squared resid <br> Log likelihood <br> Durbin-Watson stat | 0.902218 <br> 0.874281 <br> 17.38985 <br> 2116.847 <br> -40.96488 <br> 1.650804 | Mean dependent var <br> S.D. dependent var <br> Akaike info criterion <br> Schwarz criterion <br> F-statistic <br> Prob(F-statistic) |   |   | 70.3300 <br> 9.04504 <br> 792975 <br> .883751 <br> 2.29408 <br> .000292 |

将 $X_{1}=35, X_{2}=20000$ 代入回归方程, 可得

$$
Y=626.51-9.7906 \times 35+0.0286 \times 20000=856.20 \text { (元) }
$$

由于

$$
\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1}=\left(\begin{array}{rrr}
5.32536028 & -0.36302110 & 0.00053817 \\
-0.36302110 & 0.03381604 & -0.00005958 \\
0.00053817 & -0.00005958 & 0.00000011
\end{array}\right)
$$

因此, 取 $\boldsymbol{X}_{0}=\left(\begin{array}{lll}1 & 35 & 20000\end{array}\right), Y$ 均值的预测的标准差为

$$
\begin{aligned}
S_{\hat{Y}_{0}} & =\sqrt{\hat{\sigma}^{2} \boldsymbol{X}_{0}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}_{0}^{\prime}}=\sqrt{302.41 \times 4.539} \\
& =\sqrt{1372.62}=37.05
\end{aligned}
$$

在 $5 \%$ 的显著性水平下, 自由度为 $10-2-1=7$ 的 $t$ 分布的临界值为 $t_{0.025}(7)=2.365$,于是 $Y$ 均值的 $95 \%$ 的预测区间为

$$
856.20 \pm 2.365 \times 37.05 \text { 或 }(768.58,943.82)
$$

同样容易得到 $Y$ 个值的预测的标准差为

$$
\begin{aligned}
S_{\hat{Y}_{0}} & =\sqrt{\hat{\sigma}^{2}\left[1+\boldsymbol{X}_{0}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}_{0}^{\prime}\right]}=\sqrt{302.41 \times 5.539} \\
& =\sqrt{1675.03}=40.93
\end{aligned}
$$